The value of the integral `I = int_(1//2014)^(2014)(tan^(-1) x)/x dx` is

Options

A. `(pi)/4 log2014`
B. `(pi)/2 log 2014`
C. `pi log 2014`
D. `1/2 log 2014`

Answer

To solve the integral \( I = \int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1} x}{x} \, dx \), we can use the property of definite integrals and a substitution to simplify the calculation. Here’s a step-by-step breakdown of the solution:

### Step 1: Use the property of definite integrals
We can use the property that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
In our case, let \( a = \frac{1}{2014} \) and \( b = 2014 \). Thus, we have:
\[ I = \int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1} x}{x} \, dx = \int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1} \left( 2014 - x \right)}{2014 - x} \, dx \]

### Step 2: Change of variable
Let \( x = \frac{1}{t} \). Then, \( dx = -\frac{1}{t^2} dt \). The limits change as follows:
- When \( x = \frac{1}{2014} \), \( t = 2014 \)
- When \( x = 2014 \), \( t = \frac{1}{2014} \)

Thus, the integral becomes:
\[ I = \int_{2014}^{\frac{1}{2014}} \frac{\tan^{-1} \left( \frac{1}{t} \right)}{\frac{1}{t}} \left(-\frac{1}{t^2}\right) dt = \int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1} \left( \frac{1}{t} \right)}{t} dt \]

### Step 3: Use the identity for inverse tangent
We know that:
\[ \tan^{-1} \left( \frac{1}{t} \right) = \frac{\pi}{2} - \tan^{-1}(t) \]
Thus, we can rewrite the integral:
\[ I = \int_{\frac{1}{2014}}^{2014} \frac{\frac{\pi}{2} - \tan^{-1}(t)}{t} dt \]

### Step 4: Split the integral
Now, we can split the integral into two parts:
\[ I = \int_{\frac{1}{2014}}^{2014} \frac{\pi/2}{t} dt - \int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}(t)}{t} dt \]
The first integral can be computed as:
\[ \int_{\frac{1}{2014}}^{2014} \frac{\pi/2}{t} dt = \frac{\pi}{2} \left[ \ln t \right]_{\frac{1}{2014}}^{2014} = \frac{\pi}{2} \left( \ln(2014) - \ln\left(\frac{1}{2014}\right) \right) = \frac{\pi}{2} \left( \ln(2014) + \ln(2014) \right) = \pi \ln(2014) \]

### Step 5: Combine the results
Now we have:
\[ I = \pi \ln(2014) - I \]
Adding \( I \) to both sides gives:
\[ 2I = \pi \ln(2014) \]
Thus, we find:
\[ I = \frac{\pi}{2} \ln(2014) \]

### Final Answer
The value of the integral is:
\[ \boxed{\frac{\pi}{2} \ln(2014)} \]